description

analysis

• \(DP\)，这是\(Claris\)神仙的题…？

• 既然是\(2^{16}\)可以拆成两个\(2^8\)的位运算

• 照着打就行了

code

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll long long
#define fo(i,a,b) for (ll i=a;i<=b;++i)
#define fd(i,a,b) for (ll i=a;i>=b;--i)

using namespace std;

ll f[260][260][2];
char opt[10];
ll n,type,ans1,ans2;

inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
    while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int main()
{
    freopen("binary.in","r",stdin);
    freopen("binary.out","w",stdout);
    n=read(),scanf("%s",&opt),type=read();
    fo(j,1,n)
    {
        ll tmp=read(),x=tmp/256,y=tmp%256;
        if (j>1)
        {
            ans1=ans2=0;
            fo(i,0,255)
            {
                if (!f[i][y][1])continue;
                ll z;
                if (opt[0]=='a')z=i&x;
                else if (opt[0]=='o')z=i|x;
                else z=i^x;
                ll tot=(z<<8)+f[i][y][0];
                if (tot>ans1)ans1=tot,ans2=f[i][y][1];
                else if (tot==ans1)ans2+=f[i][y][1];
            }
            printf("%lld",ans1);
            if (type)printf(" %lld",ans2);
            printf("\n");
        }
        fo(i,0,255)
        {
            ll z;
            if (opt[0]=='a')z=i&y;
            else if (opt[0]=='o')z=i|y;
            else z=i^y;
            if (z>f[x][i][0])f[x][i][0]=z,f[x][i][1]=1;
            else if (z==f[x][i][0])++f[x][i][1];
        }
    }
    return 0;
}